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Math Help - find solutions to the equation

  1. #1
    Junior Member
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    find solutions to the equation

    Hi sorry am i doing this right?
    find all solutions to the equation

     x^6 + 2x^3 + 1 = 0
    let  u = x^3
    then  u^2 + 2u + 1 = 0
     = (u+1)^2 = 0
    thus  u^3 = -1
     = \sqrt{1} e^{\pi + 2\pi m}

    and then i solve for  m = 0..6
    is that correct?
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  2. #2
    Super Member bigwave's Avatar
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    Quote Originally Posted by Dgphru View Post
    Hi sorry am i doing this right?
    find all solutions to the equation

     x^6 + 2x^3 + 1 = 0
    let  u = x^3
    then  u^2 + 2u + 1 = 0
     = (u+1)^2 = 0
    thus  u^3 = -1
     = \sqrt{1} e^{\pi + 2\pi m}

    and then i solve for  m = 0..6
    is that correct?
    what is this for?
     = \sqrt{1} e^{\pi + 2\pi m}
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  3. #3
    Junior Member
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    Quote Originally Posted by bigwave View Post
    what is this for?
     = \sqrt{1} e^{\pi + 2\pi m}
    hmm i thought that that's how to solve it.. like to find the roots of the equation? like to put in into polar form? .. but honestly i have no idea what im doing
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Dgphru View Post
    Hi sorry am i doing this right?
    find all solutions to the equation

     x^6 + 2x^3 + 1 = 0
    let  u = x^3
    then  u^2 + 2u + 1 = 0
     = (u+1)^2 = 0
    thus  u^3 = -1
     = \sqrt{1} e^{\pi + 2\pi m}

    and then i solve for  m = 0..6
    is that correct?
    -1=e^{\pi(2m+1){i}}

    CB
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    -1=e^{\pi(2m+1){i}}

    CB
    thanks!!
    Last edited by Dgphru; March 18th 2010 at 03:14 AM.
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